Local theta correspondences between epipelagic supercuspidal representations

Abstract

In this paper we study the local theta correspondences between epipelagic supercupsidal representations of a type I classical dual pair $(G,G')$ over p-adic fields. We show that, besides an exceptional case, an epipelagic supercupsidal representation $\pi$ of ${\widetilde{G}}$ lifts to an epipelagic supercupsidal representation $\pi '$ of ${\widetilde{G}}'$ if and only if the epipelagic data of $\pi$ and $\pi '$ are related by the moment maps.

Appendices

Appendix 1: Description of an apartment and Proof of Lemma 8.2.1

First the fact that ${\mathsf {P}}\rightarrow {\mathsf {J}}$ splits follows from the work of McNinch [15]. For our case, the splitting could be constructed by an elementary method which we will explain below.

We retain the notation in Sect. 3.1. Let K be the maximal unramified extension in D defined in the following way:

• (i) $K := k$ if $D=k$;

• (ii) $K := D$ if D / k is an unramified extension;

• (iii) $K := k$ if D / k is a ramified extension and

• (iv) if D the quaternion algebra over k, then K is the unramified quadratic extension of k in D normalized by $\varpi _D$.

Let $\nu _D = {\nu }(\varpi _D)$. Then ${{\nu }_D}= \frac{1}{2}$ if and only if D / k is ramified or D is the quaternion algebra over k. Under this setting, $D=K$ if ${{\nu }_D}=1$ and $D = K \oplus \varpi _D K$ if ${{\nu }_D}= \frac{1}{2}$. In all cases, $\mathfrak {f}_K = \mathfrak {f}_D$.

1.1 Description of an apartment

We recall the explicit description of an apartment in ${\mathcal {B}}({\mathbf {G}},k)$ (c.f. [5, §2.9] and [3, §2-4]). Let ${\lceil r \rceil }$ denote the largest integer not greater than $r \in {\mathbb {R}}$. Let n be the dimension of a maximally isotropic subspace in V. Let $I := I^+\sqcup I^- \sqcup I^0$ where $I^+ = \left\{ 1, \ldots , n\right\}$, $I^- = -I^+$ and $I^0$ is any index set with $\dim _DV - 2n$ elements. Fix a basis $\left\{ e_i | i \in I\right\}$ of V such that

• (a) $\left\langle {e_i}, {e_j}\right\rangle _V = \left\langle {e_{-i}}, {e_{-j}}\right\rangle _V = 0$ and $\left\langle {e_i}, {e_{-j}}\right\rangle _V = \delta _{i,j}$ for all $i, j \in I^+$;

• (b) $e_i$ is anisotropic for $i\in I^0$ and $\left\langle {e_i}, {e_j}\right\rangle _V = 0$ for $i\in I^0$ and $i\ne j\in I$.

For $i\in I^0$, we can choose $e_i$ such that

• (i) $\left\langle {e_i}, {e_i}\right\rangle _V$ has valuation either 0 or ${{\nu }_D}$;

• (ii) $\left\langle {e_i}, {e_i}\right\rangle _V$ takes value either in ${\mathfrak {o}}_K$ or in $\varpi _D{\mathfrak {o}}_K$.⁴

Let ${\mathbf {S}}$ be the maximal k-split torus in ${\mathbf {G}}$ which stabilizes $e_iD$ for all $i\in I^+\sqcup I^-$ and fixes $e_j$ for all $j \in I^0$. Then the apartment ${\mathcal {A}}({\mathbf {S}},k)$ in ${\mathcal {B}}({\mathbf {G}},k)$ corresponds to the set of self-dual lattice functions which split under this basis. More precisely, if ${\mathscr {L}}$ is in the apartment, then there is a (unique) tuple of real numbers $(a_1,\ldots , a_n)\in {\mathbb {R}}^n$ such that

$$\begin{aligned} {\mathscr {L}}_r = \bigoplus _{i\in I} e_i \mathfrak {p}_D^{{\lceil (r-a_i)/{{\nu }_D} \rceil }} \end{aligned}$$

(17)

where

• (i) $a_{-i} = - a_i$ for $i\in I^+$ and

• (ii) $a_{i} = \frac{1}{2}{\nu }(\left\langle {e_i}, {e_i}\right\rangle _V)$.

In fact, $(a_1,\ldots ,a_n) \mapsto {\mathscr {L}}$ gives an identification of ${\mathbb {R}}^n$ with the apartment.

Remark

If ${\mathbf {G}}$ splits over an unramified extension of k, then the lattice function in (17) corresponds to a hyperspecial point in ${\mathcal {A}}({\mathbf {S}},k)$ if and only if $a_i/\nu _D \in b + {\mathbb {Z}}$ for all $i \in I$ where $b = 0$ or $\frac{1}{2}$.

1.2 Construction of the splitting

We let ${\mathscr {L}}$ be a lattice function as in (17) above. We consider two cases.

Case 1.:

First we assume that ${{\nu }_D}= 1$. In this case $D=K$. Let [r] denote the coset $r+{\mathbb {Z}}\in {\mathbb {Q}}/{\mathbb {Z}}$ and define

$$\begin{aligned} {V^{[r]}}:= \sum _{a_i\equiv r\pmod {{\mathbb {Z}}}} e_i K \quad \text {and}\quad {{{\mathscr {V}}}^r}:= {V^{[r]}}\cap {\mathscr {L}}_r = \sum _{a_i\equiv r\pmod {{\mathbb {Z}}}}e_i\mathfrak {p}_K^{r-a_i}. \end{aligned}$$

We make the following observations.

(a):

The restriction of the Hermitian sesquilinear form to ${V^{[r]}}$ is non-degenerate if $r\equiv 0$ or $\frac{1}{2}\pmod {{\mathbb {Z}}}$ and totally isotropic if otherwise.

(b):

For $r\in {\mathbb {R}}$, ${V^{[r]}}$ is in perfect pairing with ${V^{[-r]}}$. In particular, ${{\mathscr {V}}}^0$ and ${{\mathscr {V}}}^\frac{1}{2}$ have $\epsilon$-Hermitian sesquilinear forms $\left\langle {\;} , {\;} \right\rangle _V$ and $\left\langle {\;} , {\;} \right\rangle _V \varpi _D^{-1}$ which are defined over ${\mathfrak {o}}_K$.

Case 2.:

Now assume ${{\nu }_D}= \frac{1}{2}$. We define the K-module

$$\begin{aligned} {V^{[r]}}:= \sum _{a_i\equiv r} e_i K + \sum _{a_i+\frac{1}{2}\equiv r} e_i \varpi _D K \end{aligned}$$

and ${\mathfrak {o}}_K$-module

$$\begin{aligned} {{{\mathscr {V}}}^r}:= {V^{[r]}}\cap {\mathscr {L}}_r = \sum _{ a_i\equiv r}e_i\mathfrak {p}_K^{r-a_i} + \sum _{ a_i +\frac{1}{2}\equiv r } e_i\varpi _D \mathfrak {p}_K^{r-a_i-\frac{1}{2}}. \end{aligned}$$

We make the following observations.

(a):

The two K-subspaces ${V^{[r]}}$ and ${V^{[r+\frac{1}{2}]}}$ in V are different. However ${V^{[r]}}= {V^{[r+\frac{1}{2}]}}\varpi _D$ and ${{\mathscr {V}}}^{r+\frac{1}{2}} = {{\mathscr {V}}}^r\varpi _D$.

(b):

The restriction of the Hermitian sesquilinear form to ${V^{[r]}}$ is non-degenerate if $r \equiv 0$ or $\frac{1}{4}\pmod {\frac{1}{2}{\mathbb {Z}}}$ and totally isotropic if otherwise.

(c):

For $r\in {\mathbb {R}}$, ${V^{[r]}}$ is in perfect pairing with ${V^{[-r]}}$. In particular, there is an $\epsilon$-Hermitian sesquilinear form $\left\langle {\;} , {\;} \right\rangle _V$ and a $(-\epsilon )$-Hermitian sesquilinear form $\left\langle {\;} , {\;} \right\rangle _V \varpi _D^{-1}$ defined on ${{\mathscr {V}}}^0$ and ${{\mathscr {V}}}^{\frac{1}{4}}$ respectively. Both forms are defined over ${\mathfrak {o}}_K$.

Thanks to the definitions of ${V^{[r]}}$ and ${{{\mathscr {V}}}^r}$, the following holds for both Cases 1 and 2:

• (i) $\dim _K {V^{[r]}}= \dim _{\mathfrak {f}_D}{\mathscr {L}}_r/{\mathscr {L}}_{r^+}$ and the natural inclusion ${{{\mathscr {V}}}^r}\hookrightarrow {\mathscr {L}}_r$ induces an isomorphism

  

  (18)

• (ii) $V = \bigoplus _{[r]\in {\mathbb {Q}}/{\mathbb {Z}}} {V^{[r]}}$.

• (iii) Define an ${\mathfrak {o}}_k$-group scheme:

  $$\begin{aligned} {\mathscr {Q}}:= \mathrm{{U}}({{\mathscr {V}}}^0)\times \mathrm{{U}}({{\mathscr {V}}}^{\frac{1}{2}{{\nu }_D}})\times \prod _{r\in (0,\frac{1}{2}{{\nu }_D})} {\mathrm {GL}}_{{\mathfrak {o}}_K}({{\mathscr {V}}}^{r}). \end{aligned}$$

  Let Q and ${\mathsf {Q}}$ denote the generic fiber and special fiber of ${\mathscr {Q}}$ respectively.

• (iv) The lattices ${{{\mathscr {V}}}^r}$ give a vertex y in the building of Q. Clearly,

  $$\begin{aligned} {\mathscr {Q}}= Q_{y},\quad Q_{y,0^+} = Q_{y,1} \quad \text {and}\quad {\mathsf {Q}}= Q_{y}/Q_{y,0^+}. \end{aligned}$$

• (v) The natural action of ${\mathscr {Q}}$ on V identifies Q with $G \cap \left( \prod _{[r]\in {\mathbb {Q}}/{\mathbb {Z}}}{\mathrm {GL}}_K({V^{[r]}})\right)$ so that ${\mathscr {Q}}= Q \cap G_{{\mathscr {L}}}$.

• (vi) The natural embedding ${\mathscr {Q}}\rightarrow G_{{\mathscr {L}}}$ induces an isomorphism of $\mathfrak {f}$-groups

  

  which is compatible with (18). This follows for the fact that the both sides are isomorphic to the right hand side of (7).

For $V'$, we likewise divide into two cases and define similar notations $V'^{[r]}$, ${{\mathscr {V}}}'^r$, $Q'$, ${\mathscr {Q}}'$ etc as above.

Proof of Lemma 8.2.1

We recall ${\mathscr {B}}= {\mathscr {L}}\otimes {\mathscr {L}}'$. For $\mu \in {\mathbb {R}}$, we define $X^{[\mu ]} = \sum _{[t]+[t'] = [\mu ]} V^{[t]}\otimes _K V'^{[t']}$. Then

• (i) $W = \bigoplus _{[\mu ]\in {\mathbb {Q}}/{\mathbb {Z}}} X^{[\mu ]}$

• (ii) ${\mathscr {X}}_\mu := \sum _{t+t' = \mu } {{\mathscr {V}}}^t\otimes _{{\mathfrak {o}}_K} {{\mathscr {V}}}'^{t'}$ equals $X^{[\mu ]}\cap {\mathscr {B}}_{\mu }$.

Using the natural inclusion ${\mathscr {X}}_\mu \hookrightarrow {\mathscr {B}}_{\mu }$, we have

• (i) ${\mathscr {X}}_{\mu +1} = {\mathscr {X}}_{\mu ^+} := X_{[\mu ]}\cap {\mathscr {B}}_{\mu ^+}$.

• (ii) .

Now we recall (13) where

$$\begin{aligned} {\mathsf {Y}}:= {\mathscr {B}}_{{\frac{1}{2m}}:{\frac{1}{2m}}^+}, \quad {\mathsf {W}}:= {\mathscr {B}}_{-{\frac{1}{2m}}:{\frac{1}{2m}}^+} \quad \text {and} \quad {\mathsf {X}}:= {\mathscr {B}}_{-{\frac{1}{2m}}:-{\frac{1}{2m}}^+}. \end{aligned}$$

Let ${\mathsf {X}}' := {\mathscr {X}}_{-{\frac{1}{2m}}}/{\mathscr {X}}_{-{\frac{1}{2m}}}\mathfrak {p}_K$. Clearly by (ii).

Note that $m\ge 2$. So ${\frac{1}{m}}<1$. The inclusion ${\mathscr {X}}_{-{\frac{1}{2m}}}\hookrightarrow {\mathscr {B}}_{-{\frac{1}{2m}}}$ gives an embedding

(19)

which splits the quotient map ${\mathsf {W}}\twoheadrightarrow {\mathsf {X}}$. The embedding ${\mathscr {Q}}\times {\mathscr {Q}}' \hookrightarrow G_{\mathscr {L}}\times G'_{{\mathscr {L}}'}$ induces a splitting of ${\mathsf {P}}\twoheadrightarrow {\mathsf {J}}$:

(20)

Note that ${\mathsf {Y}}$, ${\mathsf {X}}'$ and ${\mathsf {W}}$ are natural ${\mathsf {Q}}\times {\mathsf {Q}}'$-modules and (19) is an ${\mathsf {Q}}\times {\mathsf {Q}}'$-equivariant embedding. We get a decomposition ${\mathsf {W}}= {\mathsf {X}}' \oplus {\mathsf {Y}}$ as ${\mathsf {J}}$-modules under the splitting (20). $\square$

Appendix 2: Matrix calculations

In this appendix, we prove Lemma 7.3.3 and Lemma 8.1.1.

1.1 Proof of Lemma 7.3.3

We construct below an ${\mathscr {A}}$ defined over an algebraically closed field ${\overline{k}}$ which satisfies the lemma. The lemma and the proof below is valid for any field provided $(G,G')$ is an irreducible dual pair such that G and $G'$ are both split.

There are only several cases.

• 1. $(G,G') = ({\mathrm {GL}}(n), {\mathrm {GL}}(n'))$ with $n \le n'$. We can identify

  • (a) $W = \mathrm{Mat}_{n,n'} \oplus \mathrm{Mat}_{n,n'}$,

  • (b) $(x,y)^\star = (y,-x)$ for $(x,y) \in W$,

  • (c) $M(x,y) = xy^\top \in {{\mathfrak {g}}{\mathfrak {l}}}(n)$ and

  • (d) $M(x,y) = y^\top x \in {{\mathfrak {g}}{\mathfrak {l}}}(n')$.

  We set

  $$\begin{aligned} {\mathscr {A}}= \left\{ w= (\begin{pmatrix} a&0 \end{pmatrix}, \begin{pmatrix} b&0 \end{pmatrix}) |\begin{array}{l}a = \mathrm {diag}(a_1,\ldots , a_n) \\ b = \mathrm {diag}(b_1,\ldots , b_n) \end{array} \text { with } a_i, b_i \in {\overline{k}}\right\} . \end{aligned}$$

  For any $w \in {\mathscr {A}}$, $M(w) = ab$ and $M'(w) = \begin{pmatrix} ab &{}\quad 0 \\ 0 &{}\quad 0 \end{pmatrix}$.

• 2. $(G,G')= ({\mathrm {Sp}}(2n), \mathrm{O}(2n'+1))$ with $n \le n'$. We can choose suitable bases so that $V = {\overline{k}}^{2n}$ and $V' = {\overline{k}}^{2n'+1}$ such that $\left\langle {v_1}, {v_2}\right\rangle _V = v_1^{\top } J v_2$ and $\left\langle {v'_1}, {v'_2}\right\rangle _{V'} = v'^\top _1J'v'_2$ where

  $$\begin{aligned} J = \begin{pmatrix} 0&{}\quad I_n \\ -I_n &{}\quad 0 \end{pmatrix} \quad \text {and} \quad J' = \begin{pmatrix}0 &{}\quad 0 &{}\quad I_n \\ 0 &{}\quad I_{2n'-2n+1} &{}\quad 0\\ I_n &{}\quad 0&{}\quad 0 \end{pmatrix}. \end{aligned}$$

  Now we can identify

  • (a) $W = M_{2n'+1,2n}$,

  • (b) $w^\star = J^{-1} w^{\top } J'$,

  • (c) $M(w) = w^\star w$ and

  • (d) $M'(w) = w w^\star$.

  We consider

  $$\begin{aligned} {\mathscr {A}}= \left\{ \begin{pmatrix}a&{}\quad 0 \\ 0&{}\quad 0\\ 0 &{}\quad -b \end{pmatrix}|\begin{array}{l}a = \mathrm {diag}(a_1,\ldots , a_n), \\ b = \mathrm {diag}(b_1,\ldots , b_n), \end{array},\text { with } a_i, b_i \in {\overline{k}}\right\} . \end{aligned}$$

  For any $w\in {\mathscr {A}}$, $M(w) = \begin{pmatrix} ab &{}\quad 0\\ 0&{}\quad -ab \end{pmatrix}$ and $M'(w) = \begin{pmatrix} ab &{}\quad 0&{}\quad 0 \\ 0&{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad -ab \end{pmatrix}$.

• 3. We leave the all other cases where $(G,G') = ({\mathrm {Sp}}(2n),\mathrm{O}(2n'+2))$, $(\mathrm{O}(2n),{\mathrm {Sp}}(2n'))$ or $(\mathrm{O}(2n+1),{\mathrm {Sp}}(2n'))$ where $n\le n'$ to the reader. The formulas are similar to 2. $\square$

1.2 Proof of Lemma 8.1.1

We translate everything to the left hand side of (9) and denote the images of ${\bar{w}}$, $\lambda$, $\lambda '$, ${{\mathsf {X}}_{\lambda ,\lambda '}}$, ${{\mathsf {S}}_{\lambda }}$, $\ldots$, by ${\mathring{w}}$, ${\mathring{\lambda }}$, ${\mathring{\lambda }}'$, ${\mathring{{\mathsf {X}}}_{{\mathring{\lambda }},{\mathring{\lambda }'}}}$, ${\mathring{{\mathsf {S}}}_{\mathring{\lambda }}}$, $\ldots$ respectively. We also transport implicitly the Galois actions. Then ${\mathring{\lambda }}$ and ${\mathring{\lambda }}'$ are regular semisimple elements. It is enough to prove the statements for ${\mathring{\lambda }}$ and ${\mathring{\lambda }'}$.

• (i) First we assume that ${\mathring{\lambda }}\in {\mathrm {Hom}}_{\mathfrak {f}_{{D(E)}}}({\mathsf {V}},{\mathsf {V}})$ is full rank. In this case ${\mathring{w}}\in {\mathrm {Hom}}_{\mathfrak {f}_{{D(E)}}}({\mathsf {V}},{\mathsf {V}}')$ is full rank too. By Witt’s theorem, ${\mathring{{\mathsf {M}}}}'^{-1}({\mathring{\lambda }}')$ is a single free ${\mathring{{\mathsf {G}}}}$-orbit. Let ${\mathring{w}}'\in {\mathring{{\mathsf {X}}}_{{\mathring{\lambda }},{\mathring{\lambda }'}}}$. Then there is a unique $g \in {\mathring{{\mathsf {G}}}}$ such that ${\mathring{w}}' = g \cdot {\mathring{w}}$. Clearly $g\in {\mathrm {Stab}}_{{\mathring{{\mathsf {G}}}}}({\mathring{\lambda }})$. For every $\sigma \in \mathrm {Gal}(E/k)$,

  $$\begin{aligned} g \cdot {\mathring{w}}= {\mathring{w}}' = \sigma ({\mathring{w}}') = \sigma (g)\cdot \sigma ({\mathring{w}}) = \sigma (g)\cdot {\mathring{w}}. \end{aligned}$$

  (21)

  Since ${\mathring{{\mathsf {G}}}}$ acts freely, we have $g = \sigma (g)$. Hence $g\in {\mathring{{\mathsf {S}}}_{\mathring{\lambda }}}= ({\mathrm {Stab}}_{\mathring{{\mathsf {G}}}}({\mathring{\lambda }}))^{\mathrm {Gal}(E/k)}$. This proves (i) in these cases. Now we suppose that ${\mathring{\lambda }}$ is not full rank i.e. Case (E). This only occurs for unitary groups of equal rank. We refer to the “Proof of Lemma 7.3.3” section in “Appendix 2” for the notation. In this case, ${\mathring{{\mathsf {W}}}}= M_{nn}({\overline{\mathfrak {f}}}) \oplus M_{nn}({\overline{\mathfrak {f}}})$ are two copies of n by n matrices and ${\mathring{\lambda }}$ is of rank $n-1$. There is an element in $\mathrm {Gal}(E/k)$ exchanging the two components of ${\mathring{w}}= (A,B)$, hence A and B have the same rank $n-1$. The group ${\mathring{{\mathsf {G}}}}$ is a general linear group. Let ${\mathring{{{\mathsf {S}}}{{\mathsf {L}}}}}$ be the special linear group in ${\mathring{{\mathsf {G}}}}$. Let ${\mathring{w}}'\in {\mathring{{\mathsf {X}}}_{{\mathring{\lambda }},{\mathring{\lambda }'}}}$. Let ${\mathring{{{\mathsf {S}}}{{\mathsf {L}}}}}_{{\mathring{\lambda }}} := {\mathrm {Stab}}_{{\mathring{{{\mathsf {S}}}{{\mathsf {L}}}}}}({\mathring{\lambda }})$. It is straightforward to check that ${\mathring{w}}'$ and ${\mathring{w}}$ are in the same ${\mathring{{{\mathsf {S}}}{{\mathsf {L}}}}}_{{\mathring{\lambda }}}$-orbit on which ${\mathring{{{\mathsf {S}}}{{\mathsf {L}}}}}_{{\mathring{\lambda }}}$ acts freely. Let $g\in {\mathring{{{\mathsf {S}}}{{\mathsf {L}}}}}_{{\mathring{\lambda }}}$ such that ${\mathring{w}}' = g\cdot {\mathring{w}}$. Again by (21), g is Galois invariant, i.e. $g\in {\mathring{{\mathsf {S}}}_{\mathring{\lambda }}}$. This proves (i).

• (ii) Let $w \in {\mathscr {B}}_{-{\frac{1}{2m}}}$ be a lift of ${{\bar{w}}}$. Without loss of generality, we may assume that w is of full rank. Then $\Gamma = M(w)$ and $\Gamma ' = M'(w)$ are lifts of $\lambda$ and $\lambda '$ respectively. Let ${H_{\Gamma }}$ (resp. ${H'_{\Gamma '}}$) be the stabilizer of $\Gamma$ (resp. $\Gamma '$) in G (resp. $G'$). We recall that ${H_{\Gamma }}$ is anisotropic so ${\mathcal {B}}({H_{\Gamma }}) = \left\{ x \right\}$ as shown in the proof of Proposition 2.2.2. This implies that ${H_{\Gamma }}\subseteq G_{\mathscr {L}}$ and ${H'_{\Gamma '}}\subseteq G'_{{\mathscr {L}}'}$. Using the same argument and Witt’s theorem as in (i), for every $g' \in {H'_{\Gamma '}}$ there is a unique $g \in {H_{\Gamma }}$ such that $g' w = g^{-1} w$. The map $\tilde{\alpha }:{H'_{\Gamma '}}\rightarrow {H_{\Gamma }}$ given by $g' \mapsto g$ is a surjective homomorphism. Note that ${H_{\Gamma }}$ (resp. ${H'_{\Gamma '}}$) surjects onto ${{\mathsf {S}}_{\lambda }}$ (resp. ${{\mathsf {S}}'_{\lambda '}}$) since $\Gamma$ (resp. $\Gamma '$) is a good element (c.f. [12, Corollary 2.3.5 and Lemma 2.3.6]). Then $\tilde{\alpha }$ induces a homomorphism $\alpha :{{\mathsf {S}}'_{\lambda '}}\rightarrow {{\mathsf {S}}_{\lambda }}$.

• (iii) This follows from the proofs of (i) and (ii).

• (iv) Note that $\mathrm{Im\,}{\mathring{w}}^\star = \mathrm{Im\,}{\mathring{\lambda }}\subsetneq {\mathsf {V}}$. Let $g \in {\mathring{{\mathsf {S}}}_{\mathring{\lambda }}}$. Then $g\in {\mathring{\mathbb {S}}_{{\mathring{w}}}}$ if and only if $g|_{\mathrm{Im\,}{\mathring{w}}^*} =\mathrm{id}$ if and only if $g|_{\mathrm{Im\,}{\mathring{\lambda }}} = \mathrm{id}$ if and only if $g \in {\mathring{\mathbb {S}}_{{\mathring{\lambda }}}}$, i.e. ${\mathring{\mathbb {S}}_{{\mathring{w}}}}= {\mathring{\mathbb {S}}_{{\mathring{\lambda }}}}$. Since ${\mathring{{\mathsf {X}}}_{{\mathring{\lambda }},{\mathring{\lambda }'}}}\cong {\mathring{{\mathsf {S}}}_{\mathring{\lambda }}}/{\mathring{\mathbb {S}}_{{\mathring{w}}}}$, The last assertion is clear. $\square$

Appendix 3: Lattice model and splitting

Let ${\mathrm {Sp}}(W)$ be a symplectic group of a symplectic space W. Proposition 3.3.1 follows from Lemma A.3.1 below. One may compare the lemma with [24, §4.1] and [20].

Lemma A.3.1

Let $A_1$ and $A_2$ be two self-dual lattices in W. For $i = 1,2$, let $\omega _{A_i}:{\mathrm {Sp}}(W) \rightarrow {\mathrm {Mp}}(W)$ be the section defined by the lattice model ${{\mathscr {S}}}(A_i)$ as in (4). Then

$$\begin{aligned} \omega _{A_1}(g) = \omega _{A_2}(g) \quad \forall g\in {\mathrm {Sp}}_{A_1}\cap {\mathrm {Sp}}_{A_2}. \end{aligned}$$

Proof

We have an intertwining operator $\Xi :{{\mathscr {S}}}(A_1) \rightarrow {{\mathscr {S}}}(A_2)$ given by $(\Xi \,f)(w) = \int _{A_2}\psi (\frac{1}{2}\left\langle {a}, {w}\right\rangle ) f(w+a) {\mathrm {d}a}$ between the two lattice models. This intertwining operator is unique up to scalar.

Let $g\in {\mathrm {Sp}}_{A_1}\cap {\mathrm {Sp}}_{A_2}$. Since $g:A_2 \rightarrow A_2$ is measure preserving,

$$\begin{aligned} ((\omega _{A_2}(g)\circ \Xi f)(w)= & {} \int _{A_2}\psi (\frac{1}{2}\left\langle {a}, {g^{-1}w}\right\rangle ) f(g^{-1}w +a) {\mathrm {d}a}\\= & {} \int _{A_2}\psi (\frac{1}{2}\left\langle {ga}, {w}\right\rangle ) f(g^{-1}w +a) {\mathrm {d}a}\\= & {} \int _{A_2}\psi (\frac{1}{2}\left\langle {a}, {w}\right\rangle ) f(g^{-1}w + g^{-1}a) {\mathrm {d}a}\\= & {} (\Xi \circ \omega _{A_1}(g) f)(w). \end{aligned}$$

This proves the lemma. $\square$

Let $x\in {\mathcal {B}}({\mathbf {G}},k)$. We pick any $x'\in {\mathcal {B}}({\mathbf {G}}',k)$ and let ${\mathscr {L}}$ and ${\mathscr {L}}'$ be the lattice functions corresponding to x and $x'$ respectively. Let ${\mathscr {B}}= {\mathscr {L}}\otimes {\mathscr {L}}'$ be the tensor product lattice function and A be any self-dual lattice such that ${\mathscr {B}}_{0^+} \subseteq A \subseteq {\mathscr {B}}_{0}$. We have $G_{x,0^+}$ stabilizes A, i.e. $G_{x,0^+} \subseteq {{\mathrm {Sp}}_A}$ (see also [22, §3.3.2]). As a corollary of Lemma A.3.1, the lattice models give a canonical splitting

$$\begin{aligned} \omega _+ :\bigcup _{x\in {\mathcal {B}}({\mathbf {G}},k)} G_{x,0^+} \longrightarrow {\widetilde{G}}. \end{aligned}$$